Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z + 7}{-3z - 3} \times \dfrac{-z^2 + 8z + 9}{z^2 - 13z + 36} $
First factor out any common factors. $q = \dfrac{z + 7}{-3(z + 1)} \times \dfrac{-(z^2 - 8z - 9)}{z^2 - 13z + 36} $ Then factor the quadratic expressions. $q = \dfrac {z + 7} {-3(z + 1)} \times \dfrac {-(z - 9)(z + 1)} {(z - 9)(z - 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(z + 7) \times -(z - 9)(z + 1) } {-3(z + 1) \times (z - 9)(z - 4) } $ $q = \dfrac {-(z - 9)(z + 1)(z + 7)} {-3(z - 9)(z - 4)(z + 1)} $ Notice that $(z - 9)$ and $(z + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-\cancel{(z - 9)}(z + 1)(z + 7)} {-3\cancel{(z - 9)}(z - 4)(z + 1)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac {-\cancel{(z - 9)}\cancel{(z + 1)}(z + 7)} {-3\cancel{(z - 9)}(z - 4)\cancel{(z + 1)}} $ We are dividing by $z + 1$ , so $z + 1 \neq 0$ Therefore, $z \neq -1$ $q = \dfrac {-(z + 7)} {-3(z - 4)} $ $ q = \dfrac{z + 7}{3(z - 4)}; z \neq 9; z \neq -1 $